Antonio versus Niccolò

$f(x)=|x|^{\sqrt 3}$

Antonio's 1st statement is wrong, as the function does not take every value in $R$ , it's range is just from 0 to $\infty$

Niccolo's 1st statement is correct.

Antonio's 2nd statement is also wrong, as LHS = 2 while the RHS=0

Niccolo's 2nd statement is wrong, as the function's curve meets $y=x$ at exactly 2 points.

Thus, Niccolo wins.

(This is just an example of such a function, there could be other functions that satisfy the given property)

Antonio's 1st statement is ${\color{#20A900}{\text{correct}}}$ .

$\underline{Proof}:$

Since $x^3$ runs over all the real numbers, $f$ is surjective.
Now, $f(a)=f(b)\implies f\left(f(a)\right)=f\left(f(b)\right) \implies a^3=b^3 \implies\, a=b$ , hence $f$ is injective.
From the above we get that $f$ is bijective. $_\square$

$$

Niccolò's 1st statement is ${\color{#20A900}{\text{correct}}}$ .

$\underline{Proof}:$

For all $x\in\mathbb{R}$ we have $f\left(f(x)\right)=x^3 \implies f\left(f\left(f(x)\right)\right)=f\left(x^3\right) \implies \big[f(x)\big]^3=f\left(x^3\right)$ $_\square$

$$

Antonio's 2nd statement is ${\color{#69047E}{\text{wrong}}}$ .

$\underline{Proof}:$

We will use a counterexample: $f(x) = \begin{cases} x^{\sqrt{3}} & \quad \text{if }\: x>0 \text{ and } x\ne1 \\ -(-x)^{\sqrt{3}} & \quad \text{if }\: x<0 \\ 0 & \quad \text{if }\: x=1 \\ 1 & \quad \text{if }\: x=0 \\ \end{cases}$
It is easy to confirm that this function satisfies the given property.
At the same time, $\big[f(-1)\big]^3+\big[f(1)\big]^3=(-1)^3+0^3=-1\ne1=f(0)$ $_\square$

$$

Niccolò's 2nd statement is ${\color{#20A900}{\text{correct}}}$ .

$\underline{Proof}:$

Let $x$ be the x-coordinate of a point of intersection of the graph of $f$ and the line $y=x$ .
Then $x$ satisfies the equation $f(x)=x\;$ $(1)$ .

Consequently, $f\left(f(x)\right)=f(x) \implies x^3=f(x)\;$ $(2)$ .
$(1),\; (2) \implies x^3=x \implies x=0,\; x=1,\;$ or $x=-1$ .

This means that the only possible values for the x-coordinates of the common points (if any) are $0$ , $1$ and $-1$ .

Suppose there are exactly two points of intersection. We will examine the case that these are the ones with x-coordinates $0$ and $1$ , but the same way we can work for any other combination.
Then, $\big[\; f(0)=0$ , $f(1)=1$ and $f(-1)=k\ne-1\; \big] \quad$ $(3)$ .

Thus, $f\left(f(-1)\right)=f(k) \implies (-1)^3=f(k) \implies -1=f(k) \implies f(-1)=f\left(f(k)\right) \implies k=k^3 \implies \big[\; k=0,\; k=1,\;$ or $k=-1\; \big]$ .

The latter is impossible, because of $(3)$ , since $f$ is injective and this completes the proof. $_\square$

$$

Summing up, Niccolò has no wrong statements, while Antonio has one. The winner is $\color{#D61F06}{ Niccolò }$ .